How to Calculate Area in a Graph Using Integrals?

[SwedishFred]

Calculate area D=(x,y): -1≤X≤0 0≤Y≤ X²+4x+5

I started with dA=f(x) dx
∫f(Y=x²+4x+5) [F(x) x^3/3 + 2X²+5X] higer limit 0 lower limit -1

F(0)=0
F(-1)=-3.5
F(a)-F(b) = -3,5

I don't get this ... ??
What am i missing?

Regards!

 

[Ray Vickson]

Calculate area D=(x,y): -1≤X≤0 0≤Y≤ X²+4x+5

I started with dA=f(x) dx
∫f(Y=x²+4x+5) [F(x) x^3/3 + 2X²+5X] higer limit 0 lower limit -1

F(0)=0
F(-1)=-3.5
F(a)-F(b) = -3,5

I don't get this ... ??
What am i missing?

Regards!

1) For $F(x) = \frac{1}{3}x^3 + 2 x^2 + 5x$ we have $F(-1) \neq -3.5$.
2) $\text{Area} = F(0) - F(-1)$, not the other way round.

 

[Mark44]

Also, even if F(-1) happened to be equal to -3.5 (it isn't), 0 - (-3.5) ≠ -3.5.

One more thing - questions about derivatives or integrals should be posted in the Calculus & Beyond section, not in the Precalculus section.